For $n=18,\,$ $\displaystyle H_{2^n}\gt 9.\,$ Now, the problem will be solved if we can show that $2^{18}\le 10^{10}.\,$ This is indeed so because $2^9=512\lt 10^3\,$ such that $2^{18}=(2^9)^2\lt (10^3)^2=10^6\lt 10^{10}.$

Solution 4

We set $\displaystyle I_n=\int_1^n\frac{dx}{x}=\ln (n)\,$ show visually and formally from the properties of the Riemann integration that $H_n\ge I_n.$